I got this question wrong on an exam : Name a function that is neither O(n) nor Omega(n).
After attempting to learn this stuff on my own through youtube, I'm thinking this may be a correct answer:
(n3 (1 + sin n)) is neither O(n) nor Omega(n).
Would that be accurate?
Name a function that is neither O(n) nor Omega(n)
Saying f ∈ O(g) means the quotient
f(x)/g(x)
is bounded from above for all sufficiently large x.
f ∈ Ω(g) on the other hand means the quotient
f(x)/g(x)
is bounded below away from zero for all sufficiently large x.
So to find a function that is neither O(n) nor Ω(n) means finding a function f such that the quotient
f(x)/x
becomes arbitrarily large, and arbitrarily close to zero on every interval [y, ∞).
I'm thinking this may be a correct answer: (n^3 (1 + sin n)) is neither O(n) nor Omega(n).
Let's plug it in our quotient:
(n^3*(1 + sin n))/n = n^2*(1 + sin n)
The n^2 grows to infinity, and the factor 1 + sin n is larger than 1 for roughly three out of every six n. So one every interval [y, ∞) the quotient becomes arbitrarily large. Given an arbitrary K > 0, let N_0 = y + K + 1 and N_1 the smallest of N_0 + i, i = 0, 1, ..., 4 such that sin (N_0+i) > 0. Then f(N_1)/N_1 > (y + K + 1)² > K² + K > K.
For the Ω(n) part, it's not so easy to prove, although I believe it is satisfied.
But, we can modify the function a bit, retaining the idea of multiplying a growing function with an oscillating one in such a way that the proof becomes simple.
Instead of sin n, let us choose cos (π*n), and, to offset the zeros, add a fast decreasing function to it.
f'(n) = n^3*(1 + cos (π*n) + 1/n^4)
now,
/ n^3*(2 + 1/n^4), if n is even
f'(n) = <
\ 1/n , if n is odd
and it is obvious that f' is neither bounded from above, nor from below by any positive constant multiple of n.
I would consider something like a binary search. This is both O(log N) and Ω(log N). Since Omega is defined as a lower bound, it's not allowed to exceed the function itself -- so O(log N) definitely is not Ω(N).
I think some of the comments on the deleted answer deserve some...clarification -- perhaps even outright correction. To quote from CLRS, "Ω-notation gives a lower bound for a function to within a constant factor."
Since N2 differs from N by more than a constant factor, N2 is not Ω(N).
Related
When we say a method has the time complexity of O(n^2) is it meant in the same way as in 10^2 = 100 or does it mean that the method is at max or closest to that notation? I am really confused on how to undertand Big O. I remember something called upper bound, would that mean at max?
If means that the running time is bounded above by N².
More precisely, T(N) < C.N², where C is some constant, and the inequality is true as of a certain N*.
For example, 2N²+4N+6 = O(N²), because 2N²+4N+6 < 3N² for all N>5.
Explanation
If a method f is inside O(g), with g being another function, it means that at some point (exist some n_0 such that for all n > n_0) the function f will always output a smaller value than g for that point. However, g is allowed to have an arbitrary constant k. So f(n) <= k * g(n) for all n above some n_0. So f is allowed to first be bigger, if it then starts to be smaller and keeps being smaller.
We say f is asymptotically bounded by g. Asymptotically means that we do not care how f behaves in the beginning. Only what it will do when approaching infinity. So we discard all inputs below n_0.
Illustration
An illustration would be this:
The blue function is k * g with some constant k, the red one is f. We see that f is greater at first, but then, starting at x_0, it will always be smaller than k * g. Thus f in O(g).
Definition
Mathematically, this can be expressed by
which is the usual definition of Big-O. From the explanation above, the definition should be clear. It says that from a certain n_0 on, the function f must be smaller than k * g for all inputs. k is allowed to be some constant.
Both images are taken from Wikipedia.
Examples
Here are a couple of examples to familiarize with the definition:
n is in O(n) (trivially)
n is in O(n^2) (trivially)
5n is in O(n^2) (starting from n_0 = 5)
25n^2 is in O(n^2) (taking k = 25 or greater)
2n^2 + 4n + 6 is in O(n^2) (take k = 3, starting from n_0 = 5)
Notes
Actually,O(g) is a set in the mathematical sense. It contains all functions with the above mentioned property (which are asymptotically bounded by g).
So, although some authors write f = O(g), it is actually wrong and should be f in O(g).
There are also other, similar, sets, which only differ in the direction of the bound:
Big-O: less equals <=
Small-o: less <
Big-Omega: greater equals >=
Small-omega: greater >
Theta: Big-O and Big-Omega at the same time (equals)
I am reading an example where the following are equivalent to O(N):
O(N + P), where P < N/2
O(N + log N)
Can someone explain in laymen terms how it is that the two examples above are the same thing as O(N)?
We always take the greater one in case of addition.
In both the cases N is bigger than the other part.
In first case P < N/2 < N
In second case log N < N
Hence the complexity is O(N) in both the cases.
Let f and g be two functions defined on some subset of the real numbers. One writes
f(x) = O(g(x)) as x -> infinite
if and only if there is a positive constant M such that for all sufficiently large values of x, the absolute value of f(x) is at most M multiplied by the absolute value of g(x). That is, f(x) = O(g(x)) if and only if there exists a positive real number M and a real number x0 such that
|f(x)| <= M |g(x)| for all x > x0
So in your case 1:
f(N) = N + P <= N + N/2
We could set M = 2 Then:
|f(N)| <= 3/2|N| <= 2|N| (N0 could any number)
So:
N+p = O(N)
In your second case, we could also set M=2 and N0=1 to satify that:
|N + logN| <= 2 |N| for N > 1
Big O notation usually only provides an upper bound on the growth rate of the function, wiki. Meaning for your both cases, as P < N and logN < N. So that O(N + P) = O(2N) = O(N), The same to O(N + log N) = O(2N) = O(N). Hope that can answer your question.
For the sake of understanding you can assume that O(n) represents that the complexity is of the order of n and also that O notation represents the upper bound(or the complexity in worst case). So, when I say that O(n+p) it represents that the order of n+p.
Let's assume that in worst case p = n/2, then what would be order of n+n/2? It would still be O(n), that is, linear because constants do form a part of the Big-O notation.
Similary, for O(n+logn) because logn can never be greater than n. So, overall complexity turns out to be linear.
In short
If N is a function and C is a constant:
O(N+N/2):
If C=2, then for any N>1 :
(C=2)*N > N+N/2,
2*N>3*N/2,
2> 3/2 (true)
O(N+logN):
If C=2, then for any N>2 :
(C=2)*N > N+logN,
2*N > N+logN,
2>(N+logN)/N,
2> 1 + logN/N (limit logN/N is 0),
2>1+0 (true)
Counterexample O(N^2):
No C exists such that C*N > N^2 :
C > N^2/N,
C>N (contradiction).
Boring mathematical part
I think the source of confusion is that equals sign in O(f(x))=O(N) does not mean equality! Usually if x=y then y=x. However consider O(x)=O(x^2) which is true, but reverse is false: O(x^2) != O(x)!
O(f(x)) is an upper bound of how fast a function is growing.
Upper bound is not an exact value.
If g(x)=x is an upper bound for some function f(x), then function 2*g(x) (and in general anything growing faster than g(x)) is also an upper bound for f(x).
The formal definition is: for function f(x) to be bound by some other function g(x) if you chose any constant C then, starting from some x_0 g(x) is always greater than f(x).
f(x)=N+N/2 is the same as 3*N/2=1.5*N. If we take g(x)=N and our constant C=2 then 2*g(x)=2*N is growing faster than 1.5*N:
If C=2 and x_0=1 then for any n>(x_0=1) 2*N > 1.5*N.
same applies to N+log(N):
C*N>N+log(N)
C>(N+logN)/N
C>1+log(N)/N
...take n_0=2
C>1+1/2
C>3/2=1.5
use C=2: 2*N>N+log(N) for any N>(n_0=2),
e.g.
2*3>3+log(3), 6>3+1.58=4.68
...
2*100>100+log(100), 200>100+6.64
...
Now interesting part is: no such constant exist for N & N^2. E.g. N squared grows faster than N:
C*N > N^2
C > N^2/N
C > N
obviously no single constant exists which is greater than a variable. Imagine such a constant exists C=C_0. Then starting from N=C_0+1 function N is greater than constant C, therefore such constant does not exist.
Why is this useful in computer science?
In most cases calculating exact algorithm time or space does not make sense as it would depend on hardware speed, language overhead, algorithm implementation details and many other factors.
Big O notation provides means to estimate which algorithm is better independently from real world complications. It's easy to see that O(N) is better than O(N^2) starting from some n_0 no matter which constants are there in front of two functions.
Another benefit is ability to estimate algorithm complexity by just glancing at program and using Big O properties:
for x in range(N):
sub-calc with O(C)
has complexity of O(N) and
for x in range(N):
sub-calc with O(C_0)
sub-calc with O(C_1)
still has complexity of O(N) because of "multiplication by constant rule".
for x in range(N):
sub-calc with O(N)
has complexity of O(N*N)=O(N^2) by "product rule".
for x in range(N):
sub-calc with O(C_0)
for y in range(N):
sub-calc with O(C_1)
has complexity of O(N+N)=O(2*N)=O(N) by "definition (just take C=2*C_original)".
for x in range(N):
sub-calc with O(C)
for x in range(N):
sub-calc with O(N)
has complexity of O(N^2) because "the fastest growing term determines O(f(x)) if f(x) is a sum of other functions" (see explanation in the mathematical section).
Final words
There is much more to Big-O than I can write here! For example in some real world applications and algorithms beneficial n_0 might be so big that an algorithm with worse complexity works faster on real data.
CPU cache might introduce unexpected hidden factor into otherwise asymptotically good algorithm.
Etc...
This question already has answers here:
Difference between Big-O and Little-O Notation
(5 answers)
Closed 8 years ago.
What does nb = o(an) (o is little oh) mean, intuitively? I am just beginning to self teach my self algorithms and I am having hard time interpreting such expressions every time I see one. Here, the way I understood is that for the function nb, the rate of growth is an. But this is not making sense to me regardless of being right or wrong.
f(n)=o(g(n)) means that f(n)/g(n)->0 when n->infinite.
For your problem,it should hold a>1. (n^b)/(a^n)->0 when n->infinite, since (n^b)/(sqrt(a)^n*sqrt(a)^n))=((n^b)/sqrt(a)^n) * (1/sqrt(a)^n). Let f(n)=((n^b)/sqrt(a)^n) is a function increase first and then decrease, so you can get the maximum value of max(f(n))=M, then (n^b)/(a^n) < M/(sqrt(a)^n), since a>1, sqrt(a)>1, so (sqrt(a)^n)->infinite when n->infinite. That is M/(sqrt(a)^n)->0 when n->infinite, At last, we get (n^b)/(a^n)->0 when n->infinite. That is n^b=o(a^n) by definition.
(For simplicity I'll assume that all functions always return positive values. This is the case for example for functions measuring run-time of an algorithm, as no algorithm runs in "negative" time.)
First, a recap of big-O notation, to clear up a common misunderstanding:
To say that f is O(g) means that f grows asymptotically at most as fast as g. More formally, treating both f and g as functions of a variable n, to say that f(n) is O(g(n)) means that there is a constant K, so that eventually, f(n) < K * g(n). The word "eventually" here means that there is some fixed value N (which is a function of K, f, and g), so that if n > N then f(n) < K * g(n).
For example, the function f(n) = n + 2 is O(n^2). To see why, let K = 1. Then, if n > 10, we have n + 2 < n^2, so our conditions are satisfied. A few things to note:
For n = 1, we have f(n) = 3 and g(n) = 1, so f(n) < K * g(n) actually fails. That's ok! Remember, the inequality only needs to hold eventually, and it does not matter if the inequality fails for some small finite list of n.
We used K = 1, but we didn't need to. For example, K = 2 would also have worked. The important thing is that there is some value of K which gives us the inequality we want eventually.
We saw that n + 2 is O(n^2). This might look confusing, and you might say, "Wait, isn't n + 2 actually O(n)?" The answer is yes. n + 2 is O(n), O(n^2), O(n^3), O(n/3), etc.
Little-o notation is slightly different. Big-O notation, intuitively, says that if f is O(g), then f grows asymptotically at most as fast as g. Little-o notation says that if f is o(g), then f grows asymptotically strictly slower than g.
Formally, f is o(g) if for any (let's say positive) choice of K, eventually the inequality f(n) < K * o(g) holds. So, for instance:
The function f(n) = n is not o(n). This is because, for K = 1, there is no value of n so that f(n) < K * g(n). Intuitively, f and g grow asymptotically at the same rate, so f does not grow strictly slower than g does.
The function f(n) = n is o(n^2). Why is this? Pick your favorite positive value of K. (To see the actual point, try to make K small, for example 0.001.) Imagine graphing the functions f(n) and K * g(n). One is a straight line through the origin of positive slope, and the other is a concave-up parabola through the origin. Eventually the parabola will be higher than the line, and will stay that way. (If you remember your pre-calc/calculus...)
Now we get to your actual question: let f(n) = n^b and g(n) = a^n. You asked why f is o(g).
Presumably, the author of the original statement treats a and b as constant, positive real numbers, and moreover a > 1 (if a <= 1 then the statement is false).
The statement, in Engish, is:
For any positive real number b, and any real number a > 1, the function n^b grows asymptotically strictly slower than a^n.
This is an important thing to know if you are ever going to deal with algorithmic complexity. Put simpler, one can say "polynomials grow much slower than exponential functions." It isn't immediately obvious that this is true, and is too much to write out, so here is a reference:
https://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial
Probably you will have to have some comfort with math to be able to read any proof of this fact.
Good luck!
The super high level meaning of the statement nb is o(an) is just that exponential functions like an grow much faster than polynomial functions, like nb.
The important thing to understand when looking at big O and little o notation is that they are both upper bounds. I'm guessing that's why you're confused. nb is o(an) because the growth rate of an is much bigger. You could probably find a tighter little o upper bound on nb (one where the gap between the bound and the function is smaller) but an is still valid. It's also probably worth looking at the difference between Big O and little o.
Remember that a function f is Big O of a function g if for some constant k > 0, you can eventually find a minimum value for n so that f(n) ≤ k * g(n).
A function f is little o of a function g if for any constant k > 0 you can eventually find a minimum value for n so that f(n) ≤ k * g(n).
Note that the little o requirement is harder to fulfill, meaning that if a function f is little o of a function g, it is also Big O of g, and it means the function g grows faster than if it were just Big O of g.
In your example, if b is 3 and a is 2 and we set k to 1, we can work out the minimum value for n so that nb ≤ k * an. In this case, it's between 9 and 10 since
9³ = 729 and 1 * 2⁹ = 512, which means at 9 an is not yet greater than nb
but
10³ = 1000 and 1 * 2¹⁰ = 1024, which means n is now greater than nb.
You can see graphing these functions that n will be greater than nb for any value of n > 10. At this point we've only shown that nb is Big O of n, since Big O only requires that for some value of k > 0 (we picked 1) an ≥ nb for some minimum n (in this case it's between 9 and 10)
To show that nb is little o of an, we would have to show that for any k greater than 0 you can still find a minimum value of n so that an > nb. For example, if you picked k = .5 the minimum of 10 we found earlier doesn't work, since 10³ = 1000, and .5 * 2¹⁰ = 512. But we can just keep sliding the minimum for n out further and further, the smaller you make k the bigger the minimum for n will b. Saying nb is little o of an means no matter how small you make k we will always be able to find a big enough value for n so that nb ≤ k * an
In this example: http://www.wolframalpha.com/input/?i=2%5E%281000000000%2F2%29+%3C+%283%2F2%29%5E1000000000
I noticed that those two equations are pretty similar no matter how high you go in n. Do all algorithms with a constant to the n fall in the same time complexity category? Such as 2^n, 3^n, 4^n, etc.
They are in the same category, This does not mean their complexity is the same. They are exponential running time algorithms. Obviously 2^n < 4^n
We can see 4^n/2^n = 2^2n/2^n = 2^n
This means 4^n algorithm exponential slower(2^n times) than 2^n
The same thing happens with 3^n which is 1.5^n.
But this does not mean 2^n is something far less than 4^n, It is still exponential and will not be feasible when n>50.
Note this is happening due to n is not in the base. If they were in the base like this:
4n^k vs n^k then this 2 algorithms are asymptotically the same(as long as n is relatively small than actually data size). They would be different by linear time, just like O(n) vs c * O(n)
The time complexities O(an) and O(bn) are not the same if 1 < a < b. As a quick proof, we can use the formal definition of big-O notation to show that bn ≠ O(an).
This works by contradiction. Suppose that bn = O(an) and that 1 < a < b. Then there must be some c and n0 such that for any n ≥ n0, we have that bn ≤ c · an. This means that bn / an ≤ c for any n ≥ n0. Since b > a, it should start to become clear that this is impossible - as n grows larger, bn / an = (b / a)n will get larger and larger. In particular, if we pick any n ≥ n0 such that n > logb / a c, then we will have that
(b / a)n > (b / a)log(b/a) c = c
So, if we pick n = max{n0, c + 1}, then it's not true that bn ≤ c · an, contradicting our assumption that bn = O(an).
This means, in particular, that O(2n) ≠ O(1.5n) and that O(3n) ≠ O(2n). This is why when using big-O notation, it's still necessary to specify the base of any exponents that end up getting used.
One more thing to notice - although it looks like 21000000000/2 is approximately 1.41000000000/2, notice that these are totally different numbers. The first is of the form 10108.1ish and the second of the form 10108.2ish. That might not seem like a big difference, but it's absolutely colossal. Take, for example, 10101 and 10102. This first number is 1010, which is 10 billion and takes ten digits to write out. The second is 10100, one googol, which takes 100 digits to write out. There's a huge difference between them - the first of them is close to the world population, while the second is about the total number of atoms in the universe!
Hope this helps!
(log n)^k = O(n)? For k greater or equal to 1.
My professor presented us with this statement in class, however I am not sure what it means for a function to a have a time complexity of O(n). Even stuff like n^2 = O(n^2), how can a function f(x) have a run time complexity?
As for the statement how does it equal O(n) rather than O((logn)^k)?
(log n)^k = O(n)?
Yes. The definition of big-Oh is that a function f is in O(g(n)) if there exist positive constants N and c, such that for all n > N: f(n) <= c*g(n). In this case f(n) is (log n)^k and g(n) is n, so if we insert that into the definition we get: "there exist constants N and c, such that for all n > N: (log n)^k <= c*n". This is true so (log n)^k is in O(n).
how can a function f(x) have a run time complexity
It doesn't. Nothing about big-Oh notation is specific to run-time complexity. Big-Oh is a notation to classify the growth of functions. Often the functions we're talking about measure the run-time of certain algorithms, but we can use big-Oh to talk about arbitrary functions.
f(x) = O(g(x)) means f(x) grows slower or comparably to g(x).
Technically this is interpreted as "We can find an x value, x_0, and a scale factor, M, such that this size of f(x) past x_0 is less than the scaled size of g(x)." Or in math:
|f(x)| < M |g(x)| for all x > x_0.
So for your question:
log(x)^k = O(x)? is asking : is there an x_0 and M such that
log(x)^k < M x for all x>x_0.
The existence of such M and x_0 can be done using various limit results and is relatively simple using L'Hopitals rule .. however it can be done without calculus.
The simplest proof I can come up with that doesn't rely on L'Hopitals rule uses the Taylor series
e^z = 1 + z + z^2/2 + ... = sum z^m / m!
Using z = (N! x)^(1/N) we can see that
e^(x^(1/N)) = 1 + (N! x)^(1/N) + (N! x)^(2/N)/2 + ... (N! x)^(N/N)/N! + ...
For x>0 all terms are positive so, keeping only the Nth term we get that
e^((N! x)^(1/N)) = N! x / N! + (...)
= x + (...)
> x for x > 0
Taking logarithms of both sides (since log is monotonic increasing), then raising to Nth power (also monotonic increasing since N>0)
(N! x)^(1/N) > log x for x > 0
N! x > (log x)^n for x > 0
Which is exactly the result we need, (log x)^N < M x for some M and all x > x_0, with M = N! and x_0=0